# Math Help - Convexity of a Set

1. ## Convexity of a Set

Hello,

I have the following Problem:
A={(x,y) element of R2}: [e^(3-x^2-(y-2)^2)<=1)]
Is A convex? Is it open? Is it bounded?

I dont know how to draw this set, the only thing I know that the definition of a convex set is that it is possible to join every element of the set with another element within the set with a straight line, without leaving the set.

I know that this set is not open, because it does not include all of its boundary points (because its domain has the sign <=)
I dont know if the set is bounded, I only know that the definition of bounded is, that the set could be inside of a giant circle.

Sorry for my bad English. I hope someone can give me a good explaination about how to solve problems like this.

2. ## Re: Convexity of a Set

Since $e^x$ is monotone and $e^0=1$, we have $e^{3-x^2-(y-2)^2}\le1$ iff $3-x^2-(y-2)^2\le0$, i.e., $x^2+(y-2)^2\ge3$. Does this help?

3. ## Re: Convexity of a Set

Could you explain it to me more in detail please?

4. ## Re: Convexity of a Set

It would help if you point out exactly what you don't understand. The function $e^x$ is increasing. We have $e^x < 1$ for x < 0, $e^0 = 1$ and $e^x > 1$ for x > 0. Therefore, $e^{3-x^2-(y-2)^2}\le1$ is equivalent to $3-x^2-(y-2)^2\le0$, i.e., $x^2+(y-2)^2\ge3$. Now, the locus of $x^2+(y-2)^2=3$ is a circle with center (0, 2) and radius $\sqrt{3}$. Obviously, $x^2+(y-2)^2\ge3$ consists of points that are farther from (0, 2) then $\sqrt{3}$.

5. ## Re: Convexity of a Set

Thank you for the explaination!

6. ## Re: Convexity of a Set

I gave this example another thought.

Could I solve it in another way too? For example:
A={(x,y) element of R2}: [e(3-x^2-(y-2)^2)<=1)]

Than I have ex is <=1 for x <=0

Next I test the function 3-x^2-(y-2)^2<=0 for convexity, so I calculate the Hessian which is:
Hf=-4
The Hessian is negative so it`s convex.

ex is convex but it's domain is concave, so the set is concave.

Is this definition true?

7. ## Re: Convexity of a Set

Originally Posted by infernalmich
Next I test the function 3-x^2-(y-2)^2<=0 for convexity
This is not a function, this is an inequality. Which function do you mean?

Originally Posted by infernalmich
so I calculate the Hessian which is:
Hf=-4
Can you explain how you did this?

Originally Posted by infernalmich
ex is convex but it's domain is concave
The domain of ex is the set of real numbers, which is convex. I am not even sure what a concave set is. What theorem did you use here?

Originally Posted by infernalmich
Is this definition true?
A definition cannot be true or false.

8. ## Re: Convexity of a Set

I understood that my answer was completely stupid. I was thinking totally in a wrong direction.

Now another question, if we turn the set around and say:
A={(x,y) element of R2}: [e(3-x^2-(y-2)^2)>=1)]

ex >=0

therefore
(3-x^2-(y-2)^2) >= 0

this is true for:
y=-((3-x2)1/2-1) or y=(3-x2)1/2+1

so this function is:
-closed because it's complement is open
-bounded because it can be drawn in a large enough circle
-convex because every point inside the set can be connected to each point within the set with a straight line, without leaving the set.

Is this true?
Than I finally understood the problem. If it's not true I have to push my head against the wall :-D